Find the integral of a fraction

Question

Find , in which  is some arbitrary constant.

Answer 1

∫dx/(x²-a²) = ∫dx/[(x+a)(x-a)] = ∫[A/(x+a) + B/(x-a)]dx

A/(x+a) + B/(x-a) = 1/[(x+a)(x-a)]

A(x-a) + B(x+a) = 1

Suppose x=a

A(a-a) + B(a+a) = 1

B(2a) = 1

B = 1/2a

Suppose x=-a

A(-a-a) + B(-a+a) = 1

A(-2a) = 1

A = -1/2a

Thus, ∫dx/(x²-a²) = ∫[[-1/(2a)]/(x+a) + [1/(2a)]/(x-a)]dx, which can be easily solved:

∫[[-1/(2a)]/(x+a) + [1/(2a)]/(x-a)]dx

= [-1/(2a)]ln|x+a| + [1/(2a)]ln|x-a| + C

= [1/(2a)][-ln|x+a| + ln|x-a|] + C

= [1/(2a)][ln|x-a| - ln|x+a|] + C

= [1/(2a)]ln(|x-a|/|x+a|) + C

Hope this helped!

Answer 2

∫dx/(x² - a²) = ∫1/(2a)(1/(x - a) - 1/(x + a))dx = 1/2(2a)(lnIx - aI - lnIx + aI) + C = 1/(2a)lnI(x - a)/(x + a)I + C

Answer 3

1 / (x² - a²) = 1 / [(x - a)(x + a)] = B / (x - a) + C / (x + a)

Solve for B and C and then integrate using the natural log.